Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with quantum-groups
Search options not deleted
user 22709
Questions about algebraic structures known as quantum groups, and their categories of representations. Quasitriangular Hopf algebras and their Drinfel'd twists, triangular Hopf algebras, $C^\star$ quantum groups, h-adic quantum groups, various semisimplified categories at roots of unity which are called "quantum groups", bicrossproduct quantum groups, and quantum groups coming from braided tensor categories.
8
votes
Quantum group as (relative) Drinfeld double?
Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...
start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{ …
- 1,846
0
votes
$q$-Deforming Woronowicz's Leibniz Rule
At least the Drinfel'd-Jimbo quantum groups (dual to the coordinate rings) have a structure, that is usually explored by using so-called skew-derivations satisfying exactly the rule you name (see e.g. …
- 1,846
3
votes
Non-Drinfeld–Jimbo deformations and finite quantum groups
Much depending on what you want to do with it.... ;-)
There is a duality between coordinate algebras to the Drinfel'd-Jimbo $U_q(\mathfrak{g})$ (you're title suggests you're interested rather in the …
- 1,846
3
votes
What is $\left[ \begin{array}{c} K_i;0\\ \ell\\ \end{array} \right] _{\varepsilon _i}$ in th...
You are correct and your observation is precisely the point. It is all about choices. You define the K-symbol as an element in the quantum group over $\mathbb{Q}(q)$. At a later point you more or less …
- 1,846